Bertie

Ho-ho, another riddle bites the dust.

It doesn't state in the original question that the number has to be a power of 2.

It has to be even, so let it be 2m.

Squared and with 1 subtracted, that gets you 4m^2 - 1, and that factorises as (2m - 1)(2m + 1), which is composite unless m = 1.

If you are going to do it that way, the thing to do is to square the 1.0275, square that to get 1.0275^4, square that to get 1.0275^8 and finally square again to get 1.0275^16. Then, multiply by the 1.0275^4 obtained earlier. That gets you 1.0275^20 and you can then multiply that by 8435. Can't see why you would want to do that though, unless you're a masochist or looking for some practice in multiplication.

Another possibility is to use the binomial expansion

$$(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+\dots$$

with x = 0.0275 and n = 20. The further you go into the expansion, the smaller the terms become. Keep an eye on the size of the terms to decide how far you want to go.

Call me a wimp, but I think that there are times when it's best to use  a calculator.

My dad was 83 when he died, and at no time did he use a stick.

Is this a case of an exception disproving a rule ?

Yet another variation on the cheese and biscuits question.

There are 6 choices for the first pen and then 5 choices for the second, so 6*5 = 30 combinations in all.

Sorry Melody, can't say that I like picking up on mistakes, but

$$e^{x}.e^{2}=e^{x+2}$$,    not   $$e^{2x}.$$

CPhill, when the 2 and the 1 are in the first two positions, you only have 8 numbers available for the third position and 7 for the fourth, (if the numbers are to be unique, you can't use the 2 or 1 again).

It can be done with seven weights, and I would guess that this is the optimum, but apart from it being obvious ????? I don't know how to prove it.

As Heureka points out, weights of 1 and 3 allow us to get to 4. We now need to cater for 5 and this can be done (optimally) with a weight of 9 since the 4 can be used for subtraction. Also by subtraction we can get 6,7 and 8. Moving on from 9, the 1 and 3 allow to get to 13, and the next weight we need to worry about is 14.

Using the subtraction method again, we can get 14 using a weight of 27, (27 in one pan and 1+3+9=13 in the other), and since all weights from 1 to 13 are already available, the other weights from 15 to 26 can be formed. That takes us upto 40, 1+3+9+27=40.

To get 41 a weight of 81 wiil do, (81 in one pan and 40 in the other), and the intervening weights will be covered also.

It seems fairly clear now that we are going up through the powers of 3, in which case in addition to 1, 3, 9, a 27 and 81, we need 243 and 729.

Yes, that's correct.

But you now have to shift your result back in terms of x.

You can't calculate the other angles of the triangle, because there is insufficient information to to define it.

Draw bc as a horizontal line and form the right angle at B. You have to go up (or down) from B to get to A, and you are not told how far to go. The angle at C can be anything from zero ( or just above) through to just below 90, so the angle at A can be anything from 90 to zero, approximately.